Chain rule with trig functions

All derivative rules apply when we differentiate trig functions

Let’s look at how chain rule works in combination with trigonometric functions.

Keep in mind that everything we’ve learned about power rule, product rule, and quotient rule still applies.

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Applying chain rule to the derivatives of trigonometric functions

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Using chain rule to differentiate a sine function

Example

Use chain rule to find the derivative.

$y=\sin{8x^2}$

Using substitution, we see that $u=8x^2$ and $u'=16x$.

Our original equation becomes

$y=\sin{u}$

and the derivative is

$y'=\cos{u}(u')$

Back-substituting for $u$ and $u'$, we get

$y'=\cos{8x^2}(16x)$

$y'=16x\cos{8x^2}$

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Let’s look at another example.

Keep in mind that everything we’ve learned about power rule, product rule, and quotient rule still applies.

Example

Use chain rule to find the derivative.

$y=\left(\tan{3x^4}\right)^3$

In this case we want to use a double substitution, where

$u=3x^4$

$u'=12x^3$

and

$v=\tan{u}$

$v'=\sec^2{u}(u')$

Our original equation is

$y=[v]^3$

and its derivative is

$y'=\left(3v^2\right)v'$

Back-substituting for $v$ and $v'$, we get

$y'=\left[3(\tan{u})^2\right]\left[\sec^2{u}(u')\right]$

Back-substituting again, but this time for $u$ and $u'$, we get

$y'=\left[3\left(\tan{3x^4}\right)^2\right]\left[\sec^2{3x^4}\left(12x^3\right)\right]$

$y'=36x^3\tan^2{3x^4}\sec^2{3x^4}$

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